DataFrame 열의 순서를 변경하는 방법은 무엇입니까?
다음 DataFrame( df)이 있습니다.
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.rand(10, 5))
할당으로 더 많은 열을 추가합니다.
df['mean'] = df.mean(1)
열 mean을 맨 앞으로 이동하는 방법 , 즉 다른 열의 순서는 그대로두고 첫 번째 열로 설정하는 방법은 무엇입니까?
한 가지 쉬운 방법은 필요에 따라 재 배열 된 열 목록으로 데이터 프레임을 재 할당하는 것입니다.
이것이 당신이 지금 가지고있는 것입니다 :
In [6]: df
Out[6]:
0 1 2 3 4 mean
0 0.445598 0.173835 0.343415 0.682252 0.582616 0.445543
1 0.881592 0.696942 0.702232 0.696724 0.373551 0.670208
2 0.662527 0.955193 0.131016 0.609548 0.804694 0.632596
3 0.260919 0.783467 0.593433 0.033426 0.512019 0.436653
4 0.131842 0.799367 0.182828 0.683330 0.019485 0.363371
5 0.498784 0.873495 0.383811 0.699289 0.480447 0.587165
6 0.388771 0.395757 0.745237 0.628406 0.784473 0.588529
7 0.147986 0.459451 0.310961 0.706435 0.100914 0.345149
8 0.394947 0.863494 0.585030 0.565944 0.356561 0.553195
9 0.689260 0.865243 0.136481 0.386582 0.730399 0.561593
In [7]: cols = df.columns.tolist()
In [8]: cols
Out[8]: [0L, 1L, 2L, 3L, 4L, 'mean']
cols원하는 방식으로 재정렬 하십시오. 다음은 마지막 요소를 첫 번째 위치로 이동 한 방법입니다.
In [12]: cols = cols[-1:] + cols[:-1]
In [13]: cols
Out[13]: ['mean', 0L, 1L, 2L, 3L, 4L]
그런 다음 다음과 같이 데이터 프레임을 재정렬합니다.
In [16]: df = df[cols] # OR df = df.ix[:, cols]
In [17]: df
Out[17]:
mean 0 1 2 3 4
0 0.445543 0.445598 0.173835 0.343415 0.682252 0.582616
1 0.670208 0.881592 0.696942 0.702232 0.696724 0.373551
2 0.632596 0.662527 0.955193 0.131016 0.609548 0.804694
3 0.436653 0.260919 0.783467 0.593433 0.033426 0.512019
4 0.363371 0.131842 0.799367 0.182828 0.683330 0.019485
5 0.587165 0.498784 0.873495 0.383811 0.699289 0.480447
6 0.588529 0.388771 0.395757 0.745237 0.628406 0.784473
7 0.345149 0.147986 0.459451 0.310961 0.706435 0.100914
8 0.553195 0.394947 0.863494 0.585030 0.565944 0.356561
9 0.561593 0.689260 0.865243 0.136481 0.386582 0.730399
다음과 같이 할 수도 있습니다.
df = df[['mean', '0', '1', '2', '3']]
다음을 사용하여 열 목록을 가져올 수 있습니다.
cols = list(df.columns.values)
출력은 다음을 생성합니다.
['0', '1', '2', '3', 'mean']
... 그러면 첫 번째 기능에 놓기 전에 수동으로 쉽게 재배치 할 수 있습니다.
원하는 순서대로 열 이름을 지정하십시오.
In [39]: df
Out[39]:
0 1 2 3 4 mean
0 0.172742 0.915661 0.043387 0.712833 0.190717 1
1 0.128186 0.424771 0.590779 0.771080 0.617472 1
2 0.125709 0.085894 0.989798 0.829491 0.155563 1
3 0.742578 0.104061 0.299708 0.616751 0.951802 1
4 0.721118 0.528156 0.421360 0.105886 0.322311 1
5 0.900878 0.082047 0.224656 0.195162 0.736652 1
6 0.897832 0.558108 0.318016 0.586563 0.507564 1
7 0.027178 0.375183 0.930248 0.921786 0.337060 1
8 0.763028 0.182905 0.931756 0.110675 0.423398 1
9 0.848996 0.310562 0.140873 0.304561 0.417808 1
In [40]: df = df[['mean', 4,3,2,1]]
이제 '평균'열이 앞에 나옵니다.
In [41]: df
Out[41]:
mean 4 3 2 1
0 1 0.190717 0.712833 0.043387 0.915661
1 1 0.617472 0.771080 0.590779 0.424771
2 1 0.155563 0.829491 0.989798 0.085894
3 1 0.951802 0.616751 0.299708 0.104061
4 1 0.322311 0.105886 0.421360 0.528156
5 1 0.736652 0.195162 0.224656 0.082047
6 1 0.507564 0.586563 0.318016 0.558108
7 1 0.337060 0.921786 0.930248 0.375183
8 1 0.423398 0.110675 0.931756 0.182905
9 1 0.417808 0.304561 0.140873 0.310562
어때 :
df.insert(0, 'mean', df.mean(1))
http://pandas.pydata.org/pandas-docs/stable/dsintro.html#column-selection-addition-deletion
귀하의 경우에는
df = df.reindex(columns=['mean',0,1,2,3,4])
당신이 원하는 것을 정확하게 할 것입니다.
제 경우 (일반 형식) :
df = df.reindex(columns=sorted(df.columns))
df = df.reindex(columns=(['opened'] + list([a for a in df.columns if a != 'opened']) ))
2018 년 8 월부터 :
열 이름이 너무 길어서 입력 할 수없는 경우 위치가있는 정수 목록을 통해 새 순서를 지정할 수 있습니다.
데이터:
0 1 2 3 4 mean
0 0.397312 0.361846 0.719802 0.575223 0.449205 0.500678
1 0.287256 0.522337 0.992154 0.584221 0.042739 0.485741
2 0.884812 0.464172 0.149296 0.167698 0.793634 0.491923
3 0.656891 0.500179 0.046006 0.862769 0.651065 0.543382
4 0.673702 0.223489 0.438760 0.468954 0.308509 0.422683
5 0.764020 0.093050 0.100932 0.572475 0.416471 0.389390
6 0.259181 0.248186 0.626101 0.556980 0.559413 0.449972
7 0.400591 0.075461 0.096072 0.308755 0.157078 0.207592
8 0.639745 0.368987 0.340573 0.997547 0.011892 0.471749
9 0.050582 0.714160 0.168839 0.899230 0.359690 0.438500
일반적인 예 :
new_order = [3,2,1,4,5,0]
print(df[df.columns[new_order]])
3 2 1 4 mean 0
0 0.575223 0.719802 0.361846 0.449205 0.500678 0.397312
1 0.584221 0.992154 0.522337 0.042739 0.485741 0.287256
2 0.167698 0.149296 0.464172 0.793634 0.491923 0.884812
3 0.862769 0.046006 0.500179 0.651065 0.543382 0.656891
4 0.468954 0.438760 0.223489 0.308509 0.422683 0.673702
5 0.572475 0.100932 0.093050 0.416471 0.389390 0.764020
6 0.556980 0.626101 0.248186 0.559413 0.449972 0.259181
7 0.308755 0.096072 0.075461 0.157078 0.207592 0.400591
8 0.997547 0.340573 0.368987 0.011892 0.471749 0.639745
9 0.899230 0.168839 0.714160 0.359690 0.438500 0.050582
그리고 OP 질문의 특정 경우 :
new_order = [-1,0,1,2,3,4]
df = df[df.columns[new_order]]
print(df)
mean 0 1 2 3 4
0 0.500678 0.397312 0.361846 0.719802 0.575223 0.449205
1 0.485741 0.287256 0.522337 0.992154 0.584221 0.042739
2 0.491923 0.884812 0.464172 0.149296 0.167698 0.793634
3 0.543382 0.656891 0.500179 0.046006 0.862769 0.651065
4 0.422683 0.673702 0.223489 0.438760 0.468954 0.308509
5 0.389390 0.764020 0.093050 0.100932 0.572475 0.416471
6 0.449972 0.259181 0.248186 0.626101 0.556980 0.559413
7 0.207592 0.400591 0.075461 0.096072 0.308755 0.157078
8 0.471749 0.639745 0.368987 0.340573 0.997547 0.011892
9 0.438500 0.050582 0.714160 0.168839 0.899230 0.359690
이 접근 방식의 주요 문제점은 동일한 코드를 여러 번 호출하면 매번 다른 결과가 생성되므로주의해야합니다.
원하는 순서로 새 열 목록을 만든 다음을 사용 df = df[cols]하여이 새 순서로 열을 재정렬해야합니다.
cols = ['mean'] + [col for col in df if col != 'mean']
df = df[cols]
보다 일반적인 접근 방식을 사용할 수도 있습니다. 이 예에서는 마지막 열 (-1로 표시)이 첫 번째 열로 삽입됩니다.
cols = [df.columns[-1]] + [col for col in df if col != df.columns[-1]]
df = df[cols]
You can also use this approach for reordering columns in a desired order if they are present in the DataFrame.
inserted_cols = ['a', 'b', 'c']
cols = ([col for col in inserted_cols if col in df]
+ [col for col in df if col not in inserted cols])
df = df[cols]
I ran into a similar question myself, and just wanted to add what I settled on. I liked the reindex_axis() method for changing column order. This worked:
df = df.reindex_axis(['mean'] + list(df.columns[:-1]), axis=1)
An alternate method based on the comment from @Jorge:
df = df.reindex(columns=['mean'] + list(df.columns[:-1]))
Although reindex_axis seems to be slightly faster in micro benchmarks than reindex, I think I prefer the latter for its directness.
Simply do,
df = df[['mean'] + df.columns[:-1].tolist()]
This function avoids you having to list out every variable in your dataset just to order a few of them.
def order(frame,var):
if type(var) is str:
var = [var] #let the command take a string or list
varlist =[w for w in frame.columns if w not in var]
frame = frame[var+varlist]
return frame
It takes two arguments, the first is the dataset, the second are the columns in the data set that you want to bring to the front.
So in my case I have a data set called Frame with variables A1, A2, B1, B2, Total and Date. If I want to bring Total to the front then all I have to do is:
frame = order(frame,['Total'])
If I want to bring Total and Date to the front then I do:
frame = order(frame,['Total','Date'])
EDIT:
Another useful way to use this is, if you have an unfamiliar table and you're looking with variables with a particular term in them, like VAR1, VAR2,... you may execute something like:
frame = order(frame,[v for v in frame.columns if "VAR" in v])
You could do the following (borrowing parts from Aman's answer):
cols = df.columns.tolist()
cols.insert(0, cols.pop(-1))
cols
>>>['mean', 0L, 1L, 2L, 3L, 4L]
df = df[cols]
Just type the column name you want to change, and set the index for the new location.
def change_column_order(df, col_name, index):
cols = df.columns.tolist()
cols.remove(col_name)
cols.insert(index, col_name)
return df[cols]
For your case, this would be like:
df = change_column_order(df, 'mean', 0)
Moving any column to any position:
import pandas as pd
df = pd.DataFrame({"A": [1,2,3],
"B": [2,4,8],
"C": [5,5,5]})
cols = df.columns.tolist()
column_to_move = "C"
new_position = 1
cols.insert(new_position, cols.pop(cols.index(column_to_move)))
df = df[cols]
Here's a way to move one existing column that will modify the existing data frame in place.
my_column = df.pop('column name')
df.insert(3, my_column.name, my_column)
This question has been answered before but reindex_axis is deprecated now so I would suggest to use:
df.reindex(sorted(df.columns), axis=1)
How about using "T"?
df.T.reindex(['mean',0,1,2,3,4]).T
@clocker: Your solution was very helpful for me, as I wanted to bring two columns in front from a dataframe where I do not know exactly the names of all columns, because they are generated from a pivot statement before. So, if you are in the same situation: To bring columns in front that you know the name of and then let them follow by "all the other columns", I came up with the following general solution;
df = df.reindex_axis(['Col1','Col2'] + list(df.columns.drop(['Col1','Col2'])), axis=1)
set():
A simple approach is using set(), in particular when you have a long list of columns and do not want to handle them manually:
cols = list(set(df.columns.tolist()) - set(['mean']))
cols.insert(0, 'mean')
df = df[cols]
You can use reindex which can be used for both axis:
df
# 0 1 2 3 4 mean
# 0 0.943825 0.202490 0.071908 0.452985 0.678397 0.469921
# 1 0.745569 0.103029 0.268984 0.663710 0.037813 0.363821
# 2 0.693016 0.621525 0.031589 0.956703 0.118434 0.484254
# 3 0.284922 0.527293 0.791596 0.243768 0.629102 0.495336
# 4 0.354870 0.113014 0.326395 0.656415 0.172445 0.324628
# 5 0.815584 0.532382 0.195437 0.829670 0.019001 0.478415
# 6 0.944587 0.068690 0.811771 0.006846 0.698785 0.506136
# 7 0.595077 0.437571 0.023520 0.772187 0.862554 0.538182
# 8 0.700771 0.413958 0.097996 0.355228 0.656919 0.444974
# 9 0.263138 0.906283 0.121386 0.624336 0.859904 0.555009
df.reindex(['mean', *range(5)], axis=1)
# mean 0 1 2 3 4
# 0 0.469921 0.943825 0.202490 0.071908 0.452985 0.678397
# 1 0.363821 0.745569 0.103029 0.268984 0.663710 0.037813
# 2 0.484254 0.693016 0.621525 0.031589 0.956703 0.118434
# 3 0.495336 0.284922 0.527293 0.791596 0.243768 0.629102
# 4 0.324628 0.354870 0.113014 0.326395 0.656415 0.172445
# 5 0.478415 0.815584 0.532382 0.195437 0.829670 0.019001
# 6 0.506136 0.944587 0.068690 0.811771 0.006846 0.698785
# 7 0.538182 0.595077 0.437571 0.023520 0.772187 0.862554
# 8 0.444974 0.700771 0.413958 0.097996 0.355228 0.656919
# 9 0.555009 0.263138 0.906283 0.121386 0.624336 0.859904
Here is a function to do this for any number of columns.
def mean_first(df):
ncols = df.shape[1] # Get the number of columns
index = list(range(ncols)) # Create an index to reorder the columns
index.insert(0,ncols) # This puts the last column at the front
return(df.assign(mean=df.mean(1)).iloc[:,index]) # new df with last column (mean) first
Hackiest method in the book
df.insert(0,"test",df["mean"])
df=df.drop(columns=["mean"]).rename(columns={"test":"mean"})
I believe @Aman's answer is the best if you know the location of the other column.
If you don't know the location of mean, but only have its name, you cannot resort directly to cols = cols[-1:] + cols[:-1]. Following is the next-best thing I could come up with:
meanDf = pd.DataFrame(df.pop('mean'))
# now df doesn't contain "mean" anymore. Order of join will move it to left or right:
meanDf.join(df) # has mean as first column
df.join(meanDf) # has mean as last column
I liked Shoresh's answer to use set functionality to remove columns when you don't know the location, however this didn't work for my purpose as I need to keep the original column order (which has arbitrary column labels).
I got this to work though by using IndexedSet from the boltons package.
I also needed to re-add multiple column labels, so for a more general case I used the following code:
from boltons.setutils import IndexedSet
cols = list(IndexedSet(df.columns.tolist()) - set(['mean', 'std']))
cols[0:0] =['mean', 'std']
df = df[cols]
Hope this is useful to anyone searching this thread for a general solution.
DataFrame.sort_index(axis=1) is quite clean.Check doc here. And then concat
참고URL : https://stackoverflow.com/questions/13148429/how-to-change-the-order-of-dataframe-columns
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